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Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)
(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0)
(v) x = cos t, y = sin t at 
Asked by Shail Last Modified
Answer 
Harshit Bhardwaj
(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.
On differentiating with respect to x, we get:
Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:
y − 5 = − 10(x − 0)
⇒ y − 5 = − 10x
⇒ 10x + y = 5
The slope of the normal at (0, 5) is 
Therefore, the equation of the normal at (0, 5) is given as:
(ii) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.
On differentiating with respect to x, we get:
Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:
The slope of the normal at (1, 3) is
Therefore, the equation of the normal at (1, 3) is given as:
(iii) The equation of the curve is y = x3.
On differentiating with respect to x, we get:
Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:
The slope of the normal at (1, 1) is
Therefore, the equation of the normal at (1, 1) is given as:
(iv) The equation of the curve is y = x2.
On differentiating with respect to x, we get:
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:
y − 0 = 0 (x − 0)
⇒ y = 0
The slope of the normal at (0, 0) is 
, which is not defined.
Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by
(v) The equation of the curve is x = cos t, y = sin t.
∴The slope of the tangent at
is −1.
When
Thus, the equation of the tangent to the given curve at 
is
The slope of the normal atis 
Therefore, the equation of the normal to the given curve at is
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